Help on proving $I=\int_{-\infty}^{\infty}xe^{-\pi{x^2}\left(a+x\over b+x\right)^2}dx=b-a$ $0<b<a$
$$I=\int_{-\infty}^{\infty}xe^{-\pi{x^2}\left(a+x\over b+x\right)^2}dx=b-a.$$
Applying integration by parts here is doesn't work.
$u=x$ then $du=dx$
$dv=e^{-\pi{x^2}\left(a+x\over b+x\right)}dx$ Then
$v=\int_{-\infty}^{-\infty}e^{-\pi{x^2}\left(a+x\over b+x\right)}dx=1$
See the solution of @Olivier
$$I=x-\int_{-\infty}^{-\infty}dx$$
It doesn't make any sense here. 
Can anyone provide a prove of this integral?
 A: Differentiating the equation 
$$\int_{-\infty}^{+\infty}f\left(x-\frac{a}{x-b}\right)dx=\int_{-\infty}^{+\infty} f(x)dx,\qquad a>0$$
with respect to $a$ and $b$ one obtains
$$\int_{-\infty}^{+\infty}\frac{1}{x-b}f'\left(x-\frac{a}{x-b}\right)dx=0\tag{1}
$$
$$\int_{-\infty}^{+\infty}\frac{1}{(x-b)^2}f'\left(x-\frac{a}{x-b}\right)dx=0\tag{2}
$$
Also
$$\int_{-\infty}^{+\infty}f'\left(x-\frac{a}{x-b}\right)dx=0\tag{3}
$$
Now consider 
$$I(a,b)=\int_{-\infty}^{+\infty}xf\left(x-\frac{a}{x-b}\right)dx.
$$
\begin{align}
\frac{\partial I}{\partial a}&=-\int_{-\infty}^{+\infty}\frac{x}{x-b}f'\left(x-\frac{a}{x-b}\right)dx\\
&=-\int_{-\infty}^{+\infty}\left(1+\frac{b}{x-b}\right)f'\left(x-\frac{a}{x-b}\right)dx=0. \quad  \text{(due to (1) and (3))}
\end{align}
\begin{align}
\frac{\partial I}{\partial b}&=-\int_{-\infty}^{+\infty}\frac{x}{(x-b)^2}f'\left(x-\frac{a}{x-b}\right)dx\\
&=-\int_{-\infty}^{+\infty}\left(\frac{1}{x-b}+\frac{b}{(x-b)^2}\right)f'\left(x-\frac{a}{x-b}\right)dx=0. \quad  \text{(due to (1) and (2))}
\end{align}
Thus  $$I(a,b)=\int_{-\infty}^{+\infty} xf(x)dx$$
Now for $\alpha>\beta$ one has
\begin{align}
\int_{-\infty}^{+\infty}xf\left(x\frac{\alpha +x}{\beta +x}\right){d}x&=\int_{-\infty}^{+\infty}xf\left(x-\frac{(\alpha-\beta)\beta}{x+\beta}+\alpha-\beta\right)dx\\
&=\int_{-\infty}^{+\infty}(x+\beta-\alpha)f\left(x-\frac{(\alpha-\beta)\beta}{x+2\beta-\alpha}\right)dx\\
&=I((\alpha-\beta)\beta,\alpha-2\beta)+(\beta-\alpha)\cdot \int_{-\infty}^{+\infty} f(x)dx
\end{align}
Finally

$$
\int_{-\infty}^{+\infty}xf\left(x\frac{\alpha +x}{\beta +x}\right)dx=\int_{-\infty}^{+\infty} xf(x)dx+(\beta-\alpha)\cdot \int_{-\infty}^{+\infty} f(x)dx,\quad \alpha>\beta.
$$

A: Here's a general approach to treat integrals of the form
$$
\int_{-\infty}^\infty x^{s-1}f(u)\text d x
$$
Where $s$ is a nonzero integer and for a set of positive real numbers $\{a_k\}$, a set of increasing real numbers $\{b_k\}$,
$$
u(x)=x-\sum_{k=1}^{n-1}\frac{a_k}{x-b_k}.
$$
Note that
$$
u'(x)=1+\sum_{k=1}^{n-1}\frac{a_k}{(x-b_k)^2}>0
$$
so $u(x)$ increases intervalwise-monotonically and hence has inverse in each interval $(-\infty,b_1),(b_1,b_2),\cdots, (b_{n-1},\infty)$, denoted by $x_1(u),x_2(u),\cdots, x_{n}(u)$ respectively. With the help of the graph, one easily verifies $u(x)$ maps each interval to $\mathbb R$. Split the real line into the intervals and take the inverse respectively,
$$
\begin{align}
&\int_{-\infty}^\infty x^{s-1}f(u(x))\text d x
\\ =&\int_{-\infty}^{b_1}x^{s-1}_1f(u(x_1))\text d x_1+\int_{b_1}^{b_2}x^{s-1}_2f(u(x_2))\text d x_2+\cdots+\int_{b_{n-1}}^{\infty}x^{s-1}_{n}f(u(x_{n}))\text d x_{n}
\\ =&\int_{-\infty}^\infty x^{s-1}f(u)\frac{\text d x_1}{\text d u}\text d u+\int_{-\infty}^\infty x^{s-1}f(u)\frac{\text d x_2}{\text d u}\text d u+\cdots+\int_{-\infty}^\infty x^{s-1}f(u)\frac{\text d x_{n}}{\text d u}\text d u
\\ =&\frac1s\int_{-\infty}^\infty f(u)\frac{\text d }{\text d u}\Big(x_1^s+\cdots+x_{n}^s\Big)\text d u
\\ =&\frac1s\int_{-\infty}^\infty f(u)\frac{\text d }{\text d u}p_s(\{x_k(u)\})\text d u
\end{align}
$$
Here $p_s(\{x_k(u)\})$ denotes the $s$ th power sum.
Consider the equation
$$
u(x)=x-\sum_{k=1}^{n-1}\frac{a_k}{x-b_k}=u,
$$
simple manipulation yields the algebraic equation
$$
(x-b_1)\cdots(x-b_{n-1})(x-u)-\sum_{k=1}^na_k\prod_{\substack{j=1\\j\ne k}}^{n-1}(x-b_k)=0
$$
It's easy to see that the $n$ roots of the equation are exactly the inverse functions $x_1(u),x_2(u),\cdots, x_{n}(u)$.
The elementary symmetric polynomials of the roots are the coefficients of the equation (signs considered), so are polynomials of $u$. By the known property of elementary symmetric polynomials, every symmetric polynomial, of course including the power sum, can be expressed in polynomials of them. We can conclude that $p_s(\{x_k(u)\})$ is a polynomial of $u$ and the desired integral is reduced into a linear combination of
$$
\int_{-\infty}^\infty u^{s'-1}f(u(x))\text d u, \quad s'=1,\cdots,s
$$
greatly simplified. The following are $s=1,2,3$. Values of higher orders may be much more tedious to work out.
$$
p_1=u+\sum_k b_k\\
p_2=u^2+\sum_k b_k^2+2\sum_k a_k\\
p_3=u^3+3\left(\sum_{k<j}b_kb_j+\sum_ka_k\right)u+\sum_kb_k^3+3\sum_ka_k\sum_kb_k+3\sum_ka_k\prod_{j\ne k}b_j
$$
corresponding integrals are
$$
\int_{-\infty}^\infty f(u)\text d x=\int_{-\infty}^\infty f(u)\text d u\\
\int_{-\infty}^\infty xf(u)\text d x=\int_{-\infty}^\infty uf(u)\text d u\\
\int_{-\infty}^\infty x^2f(u)\text d x=\int_{-\infty}^\infty u^2f(u)\text d u+\left(\sum_{1\le k<j\le n}b_kb_j+\sum_ka_k\right)\int_{-\infty}^\infty f(u)\text d u
$$
One may notice that the first one is the Glasser's Master Theorem.
Using the formula on the second line, with the same derivation as Nemo's, the identity holds
$$
\int_{-\infty}^{\infty}xf\left(x\frac{a +x}{b +x}\right)\text dx=\int_{-\infty}^{\infty} xf(x)\text dx+(b-a)\int_{-\infty}^{\infty} f(x)\text dx,\quad b<a
$$
and the final result follows from the Gaussian integral.
